The slopes of the common tangents to the parabola $(x - 1)^2 = 4(y - 2)$ and the ellipse $\frac{(x - 1)^2}{1} + \frac{(y - 2)^2}{2} = 1$ are $m_1$ and $m_2$. Then,$m_1^2 + m_2^2$ is equal to:

The circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ intersect at points $A$ and $B$. The equation of the common tangent with a positive slope to the circle and the hyperbola is:

For the hyperbola $H : x^{2} - y^{2} = 1$ and the ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ where $a > b > 0$,let $(1)$ the eccentricity of $E$ be the reciprocal of the eccentricity of $H$,and $(2)$ the line $y = \sqrt{\frac{5}{2}} x + K$ be a common tangent of $E$ and $H$. Then $4(a^{2} + b^{2})$ is equal to:

The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide. Then,the value of $b^2$ is

The locus of the midpoints of the chords of the hyperbola $x^{2}-y^{2}=4$,which touch the parabola $y^{2}=8x$,is:

At what angle do the curves $x^2 - y^2 = 5$ and $\frac{x^2}{18} + \frac{y^2}{8} = 1$ intersect at any common point?